Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 24

$$y-4=\frac{1}{24}(x-3)^2 .$$

The vertex is at $(3,4)$ and the directrix is $y=-2=-c$. The parabola is vertical and opens upwards; moreover, since $c$ is the distance between the vertex and the directrix, then $c=4+2=6$. Now, the parabola has the equation $$y-k=\frac{1}{4c}(x-h)^2\Longrightarrow y-4=\frac{1}{24}(x-3)^2 .$$